If the length of an open organ pipe is `33.3` cm, then the frequency of fifth overtone is `(v_(sound)=333ms^(-1))`

Options

A. 3000 Hz
B. 1500 Hz
C. 2500 Hz
D. 1250 Hz

Answer

a

Explanation

Wavelength of vibration in open organ pipe is given
by `lambda=(2L)/n,` , where, `n = 1, 2, 3, …`
Let `lambda` be the wavelength of stationary waves sta up in the
open organ pipe corresponding to `n=1.`
`lambda_(1)=(2L)/1rArr L = lambda_(1)/2`
The frequency of vibration in this mode is given by
`v_(1) = v/(lambda_(1) ) = v/(2L) rArr v_(1) = v/(2L)`
Similarly for `n=2, v_(2)=(2v)/(2L) = 2v_(1)`
` n=3, v_(3) = (3v)/(2L)=3v_(1)`
In general, the frequency of vibration in nth normal mode
of vibration in open organ pipe would be, `v_(n)=nv_(1)`
Hence, frequency of fifth overtone
`= 6xx "fundamental frequency" = 6xx v/(2L)`
`(6xx333)/(2xx33.3xx10^(-2))=3000` Hz